[Math] An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn.

combinatoricsprobability

Q: An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn. Assuming the drawing is WITH replacement, what is the probability of getting 1 red, 2 blue, and 3 green balls?

This is an exam question I got wrong. My answer was:

$\frac{{5 \choose 1}{2 \choose 2}{9 \choose 3}}{{16 \choose 6}} $

I checked other questions, such as this one, and they approached it the same way. What am I missing?

Best Answer

Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.